Logarithms question: $6+\log_3 (y) =\log_3(y)^5$

Question

Logarithms question: $6+\log_3 (y) =\log_3(y)^5$

Forms part of a simultaneous equation.

FULL QUESTION:

Solve for $x$ and $y$: \begin{cases} \log_3(xy) =5 \\[4px] \log_3(x)\log_3(y)=6 \end{cases}

Answer 1

Set $u=\log_3x$ and $v=\log_3y$. The system becomes $$ \begin{cases} u+v=5\\ uv=6 \end{cases} $$ and the solutions are quite easy to find.

It's quite unclear how you arrive to the title equation.

Answer 2

As we know that $\log_33=1$
$\therefore$ $$6\log_33=\log_3(y)^5-\log_3(y)$$ $$\log_3(3)^6=\log_3\left(\frac{y^5}{y}\right)$$ $$\log_3(3)^6=\log_3(y)^4$$ taking antilog both the side we can write as $$3^6=y^4$$ I don't know how did you find that equation but now from the first part of the full question $$\log_3(xy)=5$$ $\implies$ $$3^5=xy$$ put the value of $y$ in this then we get $$x=\frac{3^5}{3^{\frac{3}{2}}}$$ $$x=3^\frac{7}{2}$$

Answer 3

Just take $\log_3 (y) = t $ and $\log_3 (x) = s $. The equations become $$ s + t =5$$ $$ st = 6 $$ Thus, s,t are the roots of the equation $a^2-5a+6 =0$ whose obvious roots are 2,3.

Therefore, $(y=8,x=27)$ or $(x=8,y=27)$ are the solutions.

Answer 4

First, your "Full Question" is not the same as the first equation you post. Do you intend one of those "=" to be "-"? Second, is $log_3(y)^5$ untended to be $log(y^5)$? That would fit your "$5log_3(y)$" in your "Full Question". If you mean "$6+ log_3(y)= log_3(y^5)$" that is the same as $6+ log_3(y)= 5log_3(x)$ so that $4log_3(x)= 6$. Then $log_3(x^4)= 6$ and $x^4= 3^6$ so that $x= 3^{6/4}= 3^{3/2}$.